Question
Question: How do you find the center and vertices of the ellipse \[\dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{\dfr...
How do you find the center and vertices of the ellipse 4x2+41y2=1?
Solution
The standard form of equation of ellipse is a2x2+b2y2=1. And a, b are any real numbers. We can find the center and vertices using the equation and the values of a and b. To do this, first, we need to express the given equation in its standard form. The centre of the ellipse with the equation a2x2+b2y2=1 is x=0&y=0. The vertices are similar to the intercepts of the ellipse.
Complete step by step solution:
We are given the equation of the ellipse as 4x2+41y2=1. Comparing it with the standard form of the equation a2x2+b2y2=1, we get a2=4&b2=41. Taking the square root of both these equations to find the values, we get a=2&b=21.
We know that to find the center of the ellipse, we have to equate the x and y with zero. Thus, we get coordinates of centre as (0,0).
The vertices are the intercepts of the ellipse; thus, we can find them by substituting x and y to be zero separately.
Substituting x=0, we get